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Sort a linked list of 0s, 1s and 2s by changing links
  • Difficulty Level : Medium
  • Last Updated : 27 Jan, 2021

Given a linked list of 0s, 1s and 2s, sort it.
Examples:

Input  : 2->1->2->1->1->2->0->1->0
Output : 0->0->1->1->1->1->2->2->2
The sorted Array is 0, 0, 1, 1, 1, 1, 2, 2, 2.

Input : 2->1->0
Output : 0->1->2
The sorted Array is 0, 1, 2

Method 1: There is a solution discussed in below post that works by changing data of nodes. 
Sort a linked list of 0s, 1s and 2s
The above solution does not work when these values have associated data with them. 
For example, these three represent three colours and different types of objects associated with the colours and sort the objects (connected with a linked list) based on colours.
Method 2: In this post, a new solution is discussed that works by changing links.
Approach: Iterate through the linked list. Maintain 3 pointers named zero, one and two to point to current ending nodes of linked lists containing 0, 1, and 2 respectively. For every traversed node, we attach it to the end of its corresponding list. Finally, we link all three lists. To avoid many null checks, we use three dummy pointers zeroD, oneD and twoD that work as dummy headers of three lists.

C++




// CPP Program to sort a linked list 0s, 1s
// or 2s by changing links
#include <bits/stdc++.h>
 
/* Link list node */
struct Node {
    int data;
    struct Node* next;
};
 
Node* newNode(int data);
 
// Sort a linked list of 0s, 1s and 2s
// by changing pointers.
Node* sortList(Node* head)
{
    if (!head || !(head->next))
        return head;
 
    // Create three dummy nodes to point to
    // beginning of three linked lists. These
    // dummy nodes are created to avoid many
    // null checks.
    Node* zeroD = newNode(0);
    Node* oneD = newNode(0);
    Node* twoD = newNode(0);
 
    // Initialize current pointers for three
    // lists and whole list.
    Node* zero = zeroD, *one = oneD, *two = twoD;
 
    // Traverse list
    Node* curr = head;
    while (curr) {
        if (curr->data == 0) {
            zero->next = curr;
            zero = zero->next;
            curr = curr->next;
        } else if (curr->data == 1) {
            one->next = curr;
            one = one->next;
            curr = curr->next;
        } else {
            two->next = curr;
            two = two->next;
            curr = curr->next;
        }
    }
 
    // Attach three lists
    zero->next = (oneD->next)
? (oneD->next) : (twoD->next);
    one->next = twoD->next;
    two->next = NULL;
 
    // Updated head
    head = zeroD->next;
 
    // Delete dummy nodes
    delete zeroD;
    delete oneD;
    delete twoD;
 
    return head;
}
 
// Function to create and return a node
Node* newNode(int data)
{
    // allocating space
    Node* newNode = new Node;
 
    // inserting the required data
    newNode->data = data;
    newNode->next = NULL;
}
 
/* Function to print linked list */
void printList(struct Node* node)
{
    while (node != NULL) {
        printf("%d  ", node->data);
        node = node->next;
    }
    printf("\n");
}
 
/* Driver program to test above function*/
int main(void)
{
    // Creating the list 1->2->4->5
    Node* head = newNode(1);
    head->next = newNode(2);
    head->next->next = newNode(0);
    head->next->next->next = newNode(1);
 
    printf("Linked List Before Sorting\n");
    printList(head);
 
    head = sortList(head);
 
    printf("Linked List After Sorting\n");
    printList(head);
 
    return 0;
}

Java




// Java Program to sort a linked list 0s, 1s
// or 2s by changing links
public class Sort012 {
 
    // Sort a linked list of 0s, 1s and 2s
    // by changing pointers.
    public static Node sortList(Node head)
    {
        if(head==null || head.next==null)
        {
            return head;
        }
        // Create three dummy nodes to point to
        // beginning of three linked lists. These
        // dummy nodes are created to avoid many
        // null checks.
        Node zeroD = new Node(0);
        Node oneD = new Node(0);
        Node twoD = new Node(0);
 
        // Initialize current pointers for three
        // lists and whole list.
        Node zero = zeroD, one = oneD, two = twoD;
        // Traverse list
        Node curr = head;
        while (curr!=null)
        {
            if (curr.data == 0)
            {
                zero.next = curr;
                zero = zero.next;
                curr = curr.next;
            }
            else if (curr.data == 1)
            {
                one.next = curr;
                one = one.next;
                curr = curr.next;
            }
            else
            {
                two.next = curr;
                two = two.next;
                curr = curr.next;
            }
        }
        // Attach three lists
        zero.next = (oneD.next!=null)
? (oneD.next) : (twoD.next);
        one.next = twoD.next;
        two.next = null;
        // Updated head
        head = zeroD.next;
        return head;
    }
 
    // function to create and return a node
    public static Node newNode(int data)
    {
        // allocating space
        Node newNode = new Node(data);
        newNode.next = null;
        return newNode;
    }
   
    /* Function to print linked list */
    public static void printList(Node node)
    {
        while (node != null)
        {
            System.out.print(node.data+" ");
            node = node.next;
        }
    }
 
    public static void main(String args[]) {
        Node head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(0);
        head.next.next.next = new Node(1);
        System.out.println("
Linked List Before Sorting");
        printList(head);
        head = sortList(head); 
        System.out.println("
\nLinked List After Sorting");
        printList(head);
    }
}
 
class Node
{
    int data;
    Node next;
    Node(int data)
    {
        this.data=data;
    }
}
//This code is contributed by Gaurav Tiwari

Python3




# Python3 Program to sort a linked list
# 0s, 1s or 2s by changing links
import math
 
# Link list node
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None
 
#Node* newNode( data)
 
# Sort a linked list of 0s, 1s and 2s
# by changing poers.
def sortList(head):
    if (head == None or
        head.next == None):
        return head
 
    # Create three dummy nodes to point to
    # beginning of three linked lists.
    # These dummy nodes are created to
    # avoid many None checks.
    zeroD = Node(0)
    oneD = Node(0)
    twoD = Node(0)
 
    # Initialize current pointers for three
    # lists and whole list.
    zero = zeroD
    one = oneD
    two = twoD
 
    # Traverse list
    curr = head
    while (curr):
        if (curr.data == 0):
            zero.next = curr
            zero = zero.next
            curr = curr.next
        elif(curr.data == 1):
            one.next = curr
            one = one.next
            curr = curr.next
        else:
            two.next = curr
            two = two.next
            curr = curr.next
         
    # Attach three lists
    zero.next = (oneD.next) if (oneD.next ) \
                            else (twoD.next)
    one.next = twoD.next
    two.next = None
 
    # Updated head
    head = zeroD.next
 
    # Delete dummy nodes
    return head
 
# function to create and return a node
def newNode(data):
     
    # allocating space
    newNode = Node(data)
 
    # inserting the required data
    newNode.data = data
    newNode.next = None
    return newNode
 
# Function to print linked list
def prList(node):
    while (node != None):
        print(node.data, end = " ")
        node = node.next
     
# Driver Code
if __name__=='__main__':
     
    # Creating the list 1.2.4.5
    head = newNode(1)
    head.next = newNode(2)
    head.next.next = newNode(0)
    head.next.next.next = newNode(1)
 
    print("Linked List Before Sorting")
    prList(head)
 
    head = sortList(head)
 
    print("\nLinked List After Sorting")
    prList(head)
 
# This code is contributed by Srathore

C#




// C# Program to sort a linked list 0s, 1s
// or 2s by changing links
using System;
 
public class Sort012
{
 
    // Sort a linked list of 0s, 1s and 2s
    // by changing pointers.
    public static Node sortList(Node head)
    {
        if(head == null || head.next == null)
        {
            return head;
        }
        // Create three dummy nodes to point to
        // beginning of three linked lists. These
        // dummy nodes are created to avoid many
        // null checks.
        Node zeroD = new Node(0);
        Node oneD = new Node(0);
        Node twoD = new Node(0);
 
        // Initialize current pointers for three
        // lists and whole list.
        Node zero = zeroD, one = oneD, two = twoD;
        // Traverse list
        Node curr = head;
        while (curr != null)
        {
            if (curr.data == 0)
            {
                zero.next = curr;
                zero = zero.next;
                curr = curr.next;
            }
            else if (curr.data == 1)
            {
                one.next = curr;
                one = one.next;
                curr = curr.next;
            }
            else
            {
                two.next = curr;
                two = two.next;
                curr = curr.next;
            }
        }
         
        // Attach three lists
        zero.next = (oneD.next != null)
? (oneD.next) : (twoD.next);
        one.next = twoD.next;
        two.next = null;
         
        // Updated head
        head = zeroD.next;
        return head;
    }
 
    // function to create and return a node
    public static Node newNode(int data)
    {
        // allocating space
        Node newNode = new Node(data);
        newNode.next = null;
        return newNode;
    }
     
    /* Function to print linked list */
    public static void printList(Node node)
    {
        while (node != null)
        {
            Console.Write(node.data + " ");
            node = node.next;
        }
    }
 
    // Driver code
    public static void Main()
    {
        Node head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(0);
        head.next.next.next = new Node(1);
        Console.WriteLine("Linked List Before Sorting");
        printList(head);
        head = sortList(head);
        Console.WriteLine("\nLinked List After Sorting");
        printList(head);
    }
}
 
public class Node
{
    public int data;
    public Node next;
    public Node(int data)
    {
        this.data = data;
    }
}
 
// This code is contributed by PrinciRaj1992

Output : 

Linked List Before Sorting
1  2  0  1  
Linked List After Sorting
0  1  1  2  

Complexity Analysis: 

  • Time Complexity: O(n) where n is number of nodes in linked list. 
    Only one traversal of the linked list is needed.
  • Auxiliary Space: O(1). 
    As no extra space is required.

Thanks to Musarrat_123 for suggesting above solution in a comment here

This article is contributed by Bhaskar Kumar Mishra. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 




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